leetcode 973

问题

We have a list of points on the plane. Find the K closest points to the origin (0, 0).

(Here, the distance between two points on a plane is the Euclidean distance.)

You may return the answer in any order. The answer is guaranteed to be unique (except for the order that it is in.)

Example 1:

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Input: points = [[1,3],[-2,2]], K = 1
Output: [[-2,2]]
Explanation: 
The distance between (1, 3) and the origin is sqrt(10).
The distance between (-2, 2) and the origin is sqrt(8).
Since sqrt(8) < sqrt(10), (-2, 2) is closer to the origin.
We only want the closest K = 1 points from the origin, so the answer is just [[-2,2]].

Example 2:

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Input: points = [[3,3],[5,-1],[-2,4]], K = 2
Output: [[3,3],[-2,4]]
(The answer [[-2,4],[3,3]] would also be accepted.)

Note:

  1. 1 <= K <= points.length <= 10000
  2. -10000 < points[i][0] < 10000
  3. -10000 < points[i][1] < 10000

分析

和求第k大个数相同，使用快速排序的框架进行局部的快排，时间复杂度为O(logn*logn)，约为O(n).

代码

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class Solution {
public:
    inline int dis(vector<int> &a) {
        return a[0]*a[0]+a[1]*a[1];
    }
    void portionSort(vector<vector<int>> &a, int l, int r, int k) {
        if (l >= r) return;
        int i = l, j = r;
        int d = dis(a[l]);
        vector<int> t = a[i];
        while (i < j) {
            while (i<j && dis(a[j]) > d) --j;
            if (i<j) swap(a[i], a[j]), ++i;
            while (i<j && dis(a[i]) < d) ++i;
            if (i<j) swap(a[i], a[j]), --j;
        }
        a[i] = t;
        if (i==k) return;
        if (i<k)
            portionSort(a, i+1, r, k);
        else
            portionSort(a, l, i-1, k);
    }
    vector<vector<int>> kClosest(vector<vector<int>>& points, int K) {
        portionSort(points, 0, points.size()-1, K-1);
        return vector<vector<int>>(points.begin(), points.begin()+K);
    }
};